A popular pastime is to see who can push an object closest to the edge of a table without its going over. You push a 130 g object and release it 2.00 m from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 1.00 m to the floor, and lands 0.25 m from the edge of the table. If the coefficient of kinetic friction between the object and the table is 0.55, what was the object's speed (in m/s) as you released it, assuming that the local acceleration due to gravity is -9.80 m/s2?

time to fall from the table..

-1.0= - 1/2 g t^2
or t= you do it.
Now it went .25 m horizontally in t seconds

vtableedge*time=.25m
solve for vtableat edge.

Now one can look at the table top.

Vf^2=Vi^2+2ad where a is frictionforce/mass 0r
a=-mg*mu/m or -mu*g

vf^2=vi^2 -2*mu/g * 2.0
solve for vi

To find the object's speed as you released it, we can use the principle of conservation of energy. The initial mechanical energy of the object when it was released will equal its final mechanical energy when it reaches the edge of the table.

First, let's calculate the initial mechanical energy when the object was released:
Initial gravitational potential energy (U₁) = m * g * h₁
where m is the mass of the object (130 g = 0.13 kg),
g is the acceleration due to gravity (-9.80 m/s²),
and h₁ is the height of the table (1.00 m).

Initial kinetic energy (K₁) = 0.5 * m * v₁²
where v₁ is the initial velocity of the object.

The initial mechanical energy (E₁) is the sum of the initial gravitational potential and kinetic energy:
E₁ = U₁ + K₁

Next, let's calculate the final mechanical energy when the object reaches the edge of the table:
Final gravitational potential energy (U₂) = m * g * h₂
where h₂ is the height from the floor to the table surface (0.25 m).

Final kinetic energy (K₂) = 0.5 * m * v₂²
where v₂ is the velocity of the object when it reaches the edge of the table.

The final mechanical energy (E₂) is the sum of the final gravitational potential and kinetic energy:
E₂ = U₂ + K₂

Since the object slides without any loss of energy due to work done by friction, the initial mechanical energy (E₁) equals the final mechanical energy (E₂):
E₁ = E₂

Now, let's substitute the formulas and given values into the equation and solve for v₁:

U₁ + K₁ = U₂ + K₂

m * g * h₁ + 0.5 * m * v₁² = m * g * h₂ + 0.5 * m * v₂²

0.13 kg * (-9.80 m/s²) * 1.00 m + 0.5 * 0.13 kg * v₁² = 0.13 kg * (-9.80 m/s²) * 0.25 m + 0.5 * 0.13 kg * v₂²

Simplifying the equation:

-1.27404 + 0.065 v₁² = -0.31851 + 0.0325 v₂²

Since the object slid off the table, its final velocity (v₂) at the edge of the table will be zero. Thus, we can simplify the equation further:

-1.27404 + 0.065 v₁² = -0.31851 + 0.0325 * 0

-1.27404 + 0.065 v₁² = -0.31851

Now, let's isolate v₁² by moving the constants to the other side of the equation:

0.065 v₁² = -0.31851 + 1.27404

0.065 v₁² = 0.95553

Finally, solve for v₁ by dividing both sides of the equation by 0.065 and then taking the square root:

v₁ ≈ √(0.95553 / 0.065)
v₁ ≈ √14.70049
v₁ ≈ 3.834 m/s

Therefore, the object's speed as you released it was approximately 3.834 m/s.